Closure of $\bigcup\limits_1^\infty(\frac{1}{n+1}, \frac{1}{n})$ If $E:=\bigcup\limits_1^\infty(\frac{1}{n+1}, \frac{1}{n})$ in $\mathbb{R}$, then its interior is open, since a union of open sets is open, thus $E^\circ=E$, since $E^\circ$ is the largest open set contained in $E$.
Now, for the closure of $E$, one might consider $\bar{E}:=\bigcup\limits_1^\infty[\frac{1}{n+1}, \frac{1}{n}]=[\frac{1}{n+1}, \frac{1}{n}]$. However, I don't think that $E$ has a closure because, then, what will $\bar{E}\backslash E^\circ$ be? I don't think that this set has either closure or boundary.
I would appreciate some hints.
 A: $E=(0,1)\setminus \{1/n:n\in\mathbb{N}\}$. Hence its closure is $[0,1]$. (Why?)
A: Some minor points: You probably meant $\bigcup_1^{\infty}(1/(n+1),1/n)$ since $1/n$ is undefined for $n=0$.
Anyway, what does your intuition tell you about this set $E$?
Hopefully it's not too hard to see that 
$$
E=(0,1)\setminus\{1/n:n\ge2\}
$$
Once you can see this, only thing that you have to do is to prove the equality. To do that, you have to do two things:


*

*Show that $x\in E \implies x\in G:=(0,1)\setminus\{1/n:n\ge2\}$

*Show that $x\in G \implies x\in E$ (this is converse of 1)


I'll do 1. for you. It is clear that if $x\in E$ then $x\in (0,1)$ since $x$ is in one of $(1/(n+1), 1/n)$. But it's also clear that $x\neq 1/n$ for any $n\ge 2$. So in fact $x\in G$. 
For 2. note that if $x\in G$ then $x\in (0,1)$ and $x\neq 1/n$ for any $n\ge 2$. Now you just have two show that $x$ lies in at least one of the intervals of the form $(1/(n+1),1/n)$
After this, hopefully you can see what closure of $E$ is. 
A: $\bullet $Every subset $A$ of a topological space $S$ has a closure. The definition of the closure of $A$ is the intersection of all closed sets that have $A$ as a subset. That is, $\bar A=\cap F$ where $F=\{B\subset S : B$ is closed $\}.$ Note that 1:  $\;F \ne \phi$ because $S\in F$.... 2: $\;A\subset \bar A$.... 3: $\;\bar A$ is closed, because it is the common intersection of the family $F$ of closed sets.... 3: If $C$ is a closed set with$ A\subset C$ then $\bar A\subset C$.... 4: If $A_1\subset A_2$ then $\bar A_1\subset \bar A_2$.... and 5: $\;\overline {A_1\cup A_2}=\bar A_1 \cup \bar A_2.\quad$ $\bullet$Every subset $A$ of a topological space $S$ has a boundary. The definition of the boundary of $A$  is $\partial A=\bar A \cap \overline {S\backslash A}.$
